Hello!
It's really basic, but what is the perimeter of this figure? r is the radius of the circle and length of the square.
Hint: Fill in two more lines on this figure. First the radius to the other corner of the square, then complete the top of the square.
You will note that the triangle formed is an equilateral triangle, with an apex angle of 60 degrees. So we are cutting a 60 degree arc off from the circumference of the circle...
-Dan
Hello, Incendiary!
What is the perimeter of this figure?
$\displaystyle r$ is the radius of the circle and length of the square.
Label the vertices of the square: A (upper left), B (lower left), C (lower right), D (upper right).
Label the center of the circle with O.
Then: .$\displaystyle AD \,= \,BC \,= \,r.$
Hence, $\displaystyle \Delta AOD$ is equilateral: $\displaystyle \angle AOD \,= \,60^o \,=\,\frac{\pi}{3}$
The major arc AD has length: $\displaystyle \frac{5\pi}{3}r$
The three sides of the square has length: $\displaystyle 3r$
Therefore, the perimeter is: .$\displaystyle \boxed{P \;= \;\left(\frac{5\pi}{3} + 3\right)r}$
The total area is: .$\displaystyle \text{(area of square) + (area of circle) - (area of overlap)}$
The overlap is a segment of a 60° angle.
. . The area of the sector is: .$\displaystyle \frac{1}{6}\pi r^2$
. . The area of the triangle is: .$\displaystyle \frac{\sqrt{3}}{4}r^2$
Hence, the area of the segment is: . $\displaystyle \frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4}r^2$
Therefore, the area of the figure is:
. . $\displaystyle A \;=\;r^2 + \pi r^2 - \left(\frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4}r^2\right) \;=\;\boxed{\frac{1}{12}\left(10\pi + 12 + 3\sqrt{3}\right)}$